A smooth manifold $M$ may be equipped with many different geometric structures. For example:
- A Riemannian manifold is a pair $(M,g)$, where $g$ is a Riemannian metric.
- A manifold-with-connection is a pair $(M, \nabla)$, where $\nabla$ is an affine connection on $M$ (not necessarily torsion-free).
It turns out that on a Riemannian manifold $(M,g)$, there is a "best" connection to choose, the Levi-Civita connection. The Levi-Civita connection is characterized by being both (1) compatible with the metric, and (2) torsion-free.
To reiterate: once a metric $g$ is chosen, there are many connections $\nabla$ we can work with, but the Levi-Civita is the best one (in a certain precise sense).
Given an affine connection $\nabla$, any whatseover, we can define its associated torsion tensor by
$$T^\nabla(X,Y) = \nabla_XY - \nabla_YX - [X,Y].$$
We say that a connection is torsion-free iff $T^\nabla = 0$. Some connections (like the Levi-Civita connection) have this property, but others don't.
Finally, given an affine connection $\nabla$, any whatsoever, and a coordinate chart, we can talk about the associated Christoffel symbols, denoted $\Gamma^k_{ij}$. Again, they depend on both the choice of connection and the coordinate chart.
It is a fact (exercise) that $\Gamma^k_{ij} = \Gamma^k_{ji}$ in every coordinate chart if and only if $\nabla$ is torsion-free.
Well after this time i have figured it out , so we know the Chrystoffel symbols and we have the formula for the riemann tensor so we just to substitute them in the formula and there are alot of symmetries can be noticed so we really just need to calculate 2 values of the riemann tensor in certain elements of the basis of the tangent space and we have the problem resolved.
Best Answer
If you look here, you will see an equation expressing the Riemann curvature tensor in terms of the second partial derivatives of the metric and some of the Christoffel symbols, which can then be written in terms of the first derivatives of the metric tensor. This already gets some of the work out of the way, since the formula for the Riemann tensor in terms of the Christoffel symbols involves the partial derivatives of the Christoffel symbols, a step that is subsumed when using the above formulas. I do not think the resulting equation will look very nice, but if we work in normal coordinates around our point $p \in M$, then the Christoffel symbols vanish there, so we get
$$R_{iklm} = \frac{1}{2}\left(g_{im,kl} + g_{kl,im} - g_{il,km} - g_{km,il} \right)$$
This looks even nicer if we realize it as a sort of Kulkarni-Nomizu product.